ISRO EC 2025 Q.52: solution and answer

ISRO 2025 · ECQ.52 · 1 mark · MCQ

What is the value of base collector voltage V_BC for the given common base circuit shown in figure? For the transistor, assume α = 0.992.

ISRO EC 2025 question 52: PNP transistor in common-base configuration, base grounded (common rail at the bottom). — Figure from the ISRO EC 2025 paper. It is part of the question.

(A)
7.47 V
(B)
8.43 V
(C)
4.47 V
(D)
0.47 V

Answer and solution

ANSWER

$V_{B C} \approx 4.47 V$ , option (C).

Official answer: CSource: ISRO's final answer key of 15 May 2026

Emitter loop, base at ground: the emitter supply is $9 V$ with its positive end towards $R_{E}$ , so $I_{E} = \frac{V _{E E} - V _{E B}}{R _{E}} = \frac{9 - 0.7}{4 k Ω} = 2.075 mA$ .
Collector current in common base: $I_{C} = α I_{E} = 0.992 \times 2.075 mA \approx 2.058 mA$ .
The collector supply has its positive terminal on the base (ground) rail, so the end feeding $R_{C}$ sits at $- 9 V$ with respect to the base.
Collector voltage with respect to the base: $V_{C} = - 9 + I_{C} R_{C} = - 9 + 2.058 \times 2.2 \approx - 9 + 4.53 = - 4.47 V$ .
Then $V_{B C} = V_{B} - V_{C} = 0 - (- 4.47) = 4.47 V$ . For a PNP transistor a positive $V_{B C}$ means the collector-base junction is reverse biased, so the active-region assumption holds.
This matches option (c).

KEY IDEA

In the active region, find $I_{E}$ from the emitter loop, take $I_{C} = α I_{E}$ , then apply KVL around the collector loop to get the collector voltage relative to the base.

WATCH OUT FOR

Using $I_{C} = I_{E}$ , or forgetting that the collector supply polarity makes its far end $- 9 V$ for a PNP, and so computing $9 - I_{C} R_{C}$ or the wrong sign of $V_{B C}$ .

From the ISRO EC 2025 paper. Every question of this paper · The full paper (PDF, ISRO) · All 2 ISRO Biasing questions, with notes